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javascalaseqscala-java-interop
Fundhor
Fundhor

Reputation: 3577

Convert Java List to Scala Seq

I need to implement a method that returns a Scala Seq, in Java.

But I encounter this error:

java.util.ArrayList cannot be cast to scala.collection.Seq

Here is my code so far:

@Override
public Seq<String> columnNames() {
    List<String> a = new ArrayList<String>();
    a.add("john");
    a.add("mary");
    Seq<String> b = (scala.collection.Seq<String>) a;
    return b;
}

But scala.collection.JavaConverters doesn't seem to offer the possibility to convert as a Seq.

Upvotes: 50

Views: 82894

Answers (7)

Fundhor
Fundhor

Reputation: 3577

JavaConverters is what I needed to solve this.

import scala.collection.JavaConverters;

public Seq<String> convertListToSeq(List<String> inputList) {
    return JavaConverters.asScalaIteratorConverter(inputList.iterator())
                         .asScala()
                         .toSeq();
}

Upvotes: 55

Xavier Guihot
Xavier Guihot

Reputation: 61646

Starting Scala 2.13, package scala.jdk.javaapi.CollectionConverters replaces deprecated packages scala.collection.JavaConverters/JavaConversions:

import scala.jdk.javaapi.CollectionConverters;

// List<String> javaList = Arrays.asList("a", "b");
CollectionConverters.asScala(javaList).toSeq();
// Seq[String] = List(a, b)

Upvotes: 22

sapy
sapy

Reputation: 9584

import scala.collection.JavaConverters;
import scala.collection.Seq;

import java.util.ArrayList;

public class Helpers {
    public Seq<String> convertListToSeq(ArrayList<String> inputList) {
        return JavaConverters.collectionAsScalaIterableConverter(inputList).asScala().toSeq();
    }
}

Versions -

compile 'org.apache.spark:spark-core_2.11:2.3.1'
compile 'org.apache.spark:spark-sql_2.11:2.3.1'
compile group: 'commons-io', name: 'commons-io', version: '2.6'
compile "com.fasterxml.jackson.module:jackson-module-scala_2.11:2.8.8"

Upvotes: 5

Nikhil
Nikhil

Reputation: 1236

This worked for me! (Java 8, Spark 2.0.0)

import java.util.ArrayList;

import scala.collection.JavaConverters;
import scala.collection.Seq;

public class Java2Scala
{

    public Seq<String> getSeqString(ArrayList<String> list)
        {
            return JavaConverters.asScalaIterableConverter(list).asScala().toSeq();
        }

}

Upvotes: 7

loicmathieu
loicmathieu

Reputation: 5552

Up to 4 elements, you can simply use the factory method of the Seq class like this : 

Seq<String> seq1 =  new Set.Set1<>("s1").toSeq();
Seq<String> seq2 =  new Set.Set2<>("s1", "s2").toSeq();
Seq<String> seq3 =  new Set.Set3<>("s1", "s2", "s3").toSeq();
Seq<String> seq4 =  new Set.Set4<>("s1", "s2", "s3", "s4").toSeq();

Upvotes: 7

Neeleshkumar S
Neeleshkumar S

Reputation: 776

@Fundhor, the method asScalaIterableConverter was not showing up in the IDE. It may be due to a difference in the versions of Scala. I am using Scala 2.11. Instead, it showed up asScalaIteratorConverter. I made a slight change to your final snippet and it worked fine for me.

scala.collection.JavaConverters.asScalaIteratorConverter(columnNames.iterator()).asScala().toSeq() where columnNames is a java.util.List.

thanks !

Upvotes: 6

Dima
Dima

Reputation: 40500

JavaConversions should work. I think, you are looking for something like this: JavaConversions.asScalaBuffer(a).toSeq()

Upvotes: 25

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